On $\sigma$-subnormal subgroups and products of finite groups

1.   Introduction

We consider only finite groups.

The starting point for this note is the following nice connection between the subnormality of a subgroup $A$ of a group $G$ and the number of elements of the product $AB$ for any subgroup $B$ of $G$ showed by Levi in ^[1].

Theorem 1. Let $A$ be a subgroup of a group $G$. Then the following are equivalent.

1. $A$ is a subnormal subgroup of $G$.

2. $|AB|$ divides $|G|$ for every subgroup $B$ of $G$.

3. $|AP|$ divides $|G|$ for every Sylow $p$-subgroup $P$ of $G$ and all primes $p$.

This result is a consequence of the known Kegel-Wielandt conjecture proved by Kleidman ^[2] making use of the classification of finite simple groups.

Theorem 2. A subgroup $A$ of a group $G$ is subnormal in $G$ if and only if $A \cap P$ is a Sylow $p$-subgroup of $A$ for each Sylow $p$-subgroup $P$ of $G$ and each prime $p$.

Let $\sigma = \{ {\sigma}_{i} : i \in I \}$ be a partition of the set $\mathbb{P}$ of all prime numbers. Following Skiba ^[3,4,5], a subgroup $A$ of a finite group $G$ is said to be $\sigma$-subnormal in $G$ if $A$ can be joined to $G$ by a chain of subgroups

such that either $A_{i-1}$ normal in $A_{i}$ or $A_{i}/\text{Core}_{A_{i}}(A_{i-1})$ is a ${ \sigma}_{j}$-group for some $j \in I$, for every $1 \leq i \leq n$.

It is abundantly clear that the embedding property of $\sigma$-subnormality coincides with the subnormality when $\sigma$ is the partition of $\mathbb{P}$ into sets containing exactly one prime each.

A group $G \neq 1$ is called $\sigma$-primary if all the primes dividing $|G|$ belong to the same member of the partition $\sigma$. We stipulate that the trivial group is $\sigma$-primary.

Definition 1. A group $G$ is called $\sigma$-soluble if all chief factors of $G$ are $\sigma$-primary. $G$ is called $\sigma$-nilpotent if it is a direct product of $\sigma$-primary groups.

If $\pi = \{p_1, \cdots, p_r\}$, and $\sigma = \{\{p_1\}, \cdots, \{p_r\}, \pi^{\prime }\}$, then the class of all $\sigma$-soluble groups is just the class of all $\pi$-soluble groups, and the class of all $\sigma$-nilpotent groups is just the class of all groups having a normal Hall $\pi^{\prime }$-subgroup and a normal Sylow $p_i$ -subgroup, for all $i$. In particular, soluble and nilpotent groups are exactly the $\sigma$-soluble and $\sigma$-nilpotent groups for the partition $\sigma = \{\{2\}, \{3\}, \{5\}, ... \}$.

Skiba ^[6,7] proved that $\sigma$-soluble groups have a nice arithmetic structure.

Theorem 3. Assume that $G$ is a $\sigma$-soluble group. Then $G$ has a Hall ${\sigma}_{i}$-subgroup $E$ and every ${\sigma}_{i}$-subgroup is contained in a conjugate of $E$ for all $i \in I$. In particular, the Hall ${\sigma}_{i}$-subgroups are conjugate for all $i\in I$. Furthermore, $G$ has Hall ${\sigma}_{i}'$-subgroups.

A non-$\sigma$-nilpotent group has a non-trivial proper $\sigma$-subnormal subgroup if and only if it si not simple Therefore criteria for the $\sigma$-subnormality of a subgroup is important in the study of the normal structure of a group ^[8]. The significance of the $\sigma$-subnormal subgroups in $\sigma$-soluble groups is also apparent since they are precisely the $K-{\cal N}_{\sigma}$-subnormal subgroups. In particular, they form a distinguished sublattice of the subgroup lattice of $G$ ^[9].

It is worth mentioning that $\sigma$-subnormality has been recently studied in the locally finite case by Ferrara and Trombetti in ^[10].

Definition 2. Let $A$ be a subgroup of a $\sigma$-soluble group $G$. We say that $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$ if $|AB|$ divides $|G|$ for every Hall ${\sigma}_{i}$-subgroup $B$ of $G$.

Taking the close relationship between $\sigma$-subnormal subgroups and Hall subgroups of $\sigma$-soluble groups into account, it seems natural to think about an extension of Theorem 1 in the $\sigma$-soluble universe. Then main result here is the following $\sigma$-subnormality criterion.

Theorem A. Let $A$ be a subgroup of a $\sigma$-soluble group $G$. Then the following are equivalent.

1. $A$ is $\sigma$-subnormal in $G$.

2. $A$ satisfies $\mathcal C_{\sigma_{i}}$ in $G$ for all $i \in I$.

2.   Proof of Theorem A

The proof of Theorem A depends on the following lemma.

Lemma 1 (^[4]). Let $A$, $B$ and $N$ be subgroups of a group $G$. Suppose that $A$ is $\sigma$-subnormal in $G$ and $N$ is normal in $G$. Then:

1. $A \cap B$ is a $\sigma$-subnormal subgroup of $B$.

2. If $B$ is $\sigma$-subnormal in $A$, then $B$ is $\sigma$ -subnormal in $G$.

3. If $B$ is a $\sigma$-subnormal subgroup of $G$, then $A \cap B$ is $\sigma$ -subnormal in $G$.

4. $AN/N$ is $\sigma$-subnormal in $G/N$.

5. If $N \subseteq B$ and $B/N$ is a $\sigma$-subnormal subgroup of $G/N$, then $B$ is $\sigma$-subnormal in $G$.

6. If $L \leq B$ and $B$ is a $\sigma$-nilpotent group, then $L$ is $\sigma$ -subnormal in $B$.

7. If the primes dividing $|G:A|$ belong to $\sigma_i$, then $O^{\sigma_i}(A) = O^{\sigma_i}(G)$.

Proof of Theorem A. Assume that $A$ is $\sigma$-subnormal in $G$, and let $B$ be a Hall ${\sigma}_{i}$-subgroup of $G$ for some $i \in I$. We show that $|AB|$ divides $|G|$ by induction on the order of $G$. If $A$ is normal in $G$, then $AB$ is a subgroup of $G$ and the result follows. Suppose that $A$ is a maximal subgroup of $G$. Then $G/\text{Core}_G(A)$ is a ${\sigma}_{j}$-group for some $j \in I$. If $i \neq j$, then $B$ is contained in $\text{Core}_G(A)$, $AB = A$ and $|AB| = |A|$ divides $|G|$. If $i = j$, then $G = \text{Core}_G(A)B$ and the result also follows.

Assume that $A$ is not a maximal subgroup of $G$, and let $M$ be a $\sigma$-subnormal maximal subgroup of $G$ containing $A$. Then $A$ is $\sigma$-subnormal in $M$. By the above argument, $B \leq M$ or $G = \text{Core}_G(M)B$. In both cases, $B \cap M$ is a Hall ${\sigma}_{i}$-subgroup of $M$. By induction, $|A(B \cap M)|$ divides $|M|$. Then there exists a positive integer $a$ such that

If $B \leq M$, then $|AB|$ divides $|M|$ and the result follows. Assume that $G = \text{Core}_G(M)B = MB$. Then

Therefore the condition is necessary.

Conversely, assume that $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$ for all $i \in I$, but $A$ is not $\sigma$-subnormal in $G$. We argue by induction on the order of $G$. Let $N$ be a minimal normal subgroup of $G$. Since $G$ is $\sigma$-soluble, it follows that $N$ is a ${\sigma}_{j}$-group for some $j \in I$. If $T/N$ is a Hall ${\sigma}_{i}$-subgroup of $G/N$ for some $i \in I$, then there exists a Hall ${\sigma}_{i}$-subgroup $B$ of $G$ such that $T/N = BN/N$ by Theorem 3. Moreover, either $N \leq B$ or $N \cap B = 1$. Since $|AB|$ divides $|G| = |B||G: B|$, we conclude that $|A : A \cap B|$ divides $|G : B|$ and then $A \cap B$ is a Hall ${\sigma}_{i}$-subgroup of $A$. Hence if $N \cap B = 1$, then $AN \cap B = A \cap B$. Thus $|(AN/N)(T/N)|$ divides $|G/N|$ and $AN/N$ satisfies the $\mathcal C_{\sigma_{i}}$ in $G/N$ for all $i \in I$. By induction, $AN/N$ is $\sigma$-subnormal in $G/N$. From Lemma 1(5), we have that $AN$ is $\sigma$-subnormal in $G$. Suppose that $AN$ is a proper subgroup of $G$. Let $C$ be a Hall ${\sigma}_{i}$-subgroup of $AN$. If $i = j$, then $N$ is contained in $C$ and $C = (A \cap C)N$. In this case, $AC = AN$ and so $|AC|$ divides $|AN|$. Suppose that $i \neq j$. By Theorem 3, there exists a Hall ${\sigma}_{i}$-subgroup $B$ of $G$ such that $C \leq B$. Since $|AB|$ divides $|G| = |B||G: B|$, it follows that $|A : A \cap B|$ divides $|G : B|$ and so $A \cap B$ is a Hall ${\sigma}_{i}$-subgroup of $A$. Hence $C = A \cap B$ and $AC = A$. In particular, $|AC|$ divides $|AN|$. Consequently, $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $AN$ for all $i \in I$. Then the induction hypothesis again applies and gives that $A$ is $\sigma$-subnormal in $AN$. From Lemma 1(2), we conclude that $A$ is $\sigma$-subnormal in $G$. Therefore we may assume that $G = AN$. From Lemma 1(7), we conclude that $O^{\sigma_i}(A) = O^{\sigma_i}(G)$. This yields $O^{\sigma_i}(G) \leq \text{Core}_G(A)$. If $O^{\sigma_i}(G) \neq 1$, we can take $N \leq O^{\sigma_i}(G)$ and conclude $A = AN$ is $\sigma$-subnormal in $G$ and if $O^{\sigma_i}(G) = 1$, then $G$ is a $\sigma_i$-group and then $A$ is obviously a $\sigma$-subnormal subgroup of $G$, as desired.

3.   Corollaries

We now derive some consequences of Theorem A, the first being a particular case of the theorem.

Corollary 1. Let $A$ be a $\sigma_i$-subgroup of a $\sigma$-soluble group $G$. Then $A$ is $\sigma$-subnormal in $G$ if and only if $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$.

Proof. Only the necessity of the condition is in doubt. Assume that $B$ is a Hall $\sigma_i$-subgroup of $G$. Since $|AB|$ divides $|G|$, it follows that every prime dividing $|AB|$ belongs to $\sigma_i$. Therefore, $AB = B$ and so $A \leq B$. Since the Hall $\sigma_i$-subgroups are conjugate, we have that $A \leq O_{\sigma_i}(G)$ and so $A$ is $\sigma$-subnormal in $O_{\sigma_i}(G)$. Since $O_{\sigma_i}(G)$ is normal in $G$, we have that $A$ is $\sigma$-subnormal in $G$ by Lemma 1(2).

A nice consequence of Theorem A is the following extension of a classical result of Kegel due to Skiba ^[6].

Corollary 2. A subgroup $A$ of a $\sigma$-soluble group $G$ is $\sigma$-subnormal in $G$ if and only if $A \cap B$ is a Hall ${\sigma}_{i}$-subgroup of $A$ for every Hall ${\sigma}_{i}$-subgroup $B$ of $G$ for all $i \in I$.

Proof. Assume that $A$ is $\sigma$-subnormal in $G$ and let $B$ be a Hall ${\sigma}_{i}$-subgroup of $G$. Then $|AB|$ divides $|G|$ and so $|A : A \cap B|$ is a $\sigma_{i}'$-number. Hence $A \cap B$ is a Hall ${\sigma}_{i}$-subgroup of $A$.

Conversely, if $B$ is a Hall ${\sigma}_{i}$-subgroup of $G$ such that $A \cap B$ is a Hall ${\sigma}_{i}$-subgroup of $A$, then $|A : A \cap B|$ divides the order of a Hall $\sigma_{i}'$-subgroup of $G$. Consequently, $|AB|$ divides $|G|$ and hence $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$ for all $i \in I$. By Theorem A, $A$ is $\sigma$-subnormal in $G$.

It is clear that the class $\mathcal{N}_{\sigma }$ of all $\sigma$-nilpotent groups behaves in the class of all $\sigma$-soluble groups like nilpotent groups in the class of all soluble groups. In fact, $\mathcal{N}_{\sigma }$ is a subgroup-closed saturated Fitting formation ^[4].

The $\mathcal{N}_{\sigma }$-radical of a group $G$ is called the $\sigma$-Fitting subgroup of $G$ and it is denoted by $F_{\sigma}(G)$. By ^[9], $F_{\sigma}(G)$ contains every $\sigma$-subnormal $\sigma$-nilpotent subgroup of $G$. Consequently, a group $G$ is $\sigma$-nilpotent if and only if every subgroup of $G$ is $\sigma$-subnormal in $G$. Therefore the following $\sigma$-version of ^[1] holds.

Corollary 3. Let $G$ be a $\sigma$-soluble group. Then $G$ is $\sigma$-nilpotent if and only if every $\sigma_i$-subgroup of $G$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$ for all $i \in I$.

Proof. If $G$ is $\sigma$-nilpotent and $i \in I$, then every $\sigma_i$-subgroup $A$ of $G$ is $\sigma$-subnormal in $G$. By Theorem A, $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$. Conversely, let $i \in I$ and let $A$ be a Hall $\sigma_i$-subgroup of $G$. Then $|AB|$ divides $|G|$ for every Hall $\sigma_i$-subgroup $B$ of $G$. Since $|AB|$ is a $\sigma_i$-number, it follows that $A = B$. Therefore $G$ has a normal Hall $\sigma_i$-subgroup for all $i \in I$ and hence $G$ is $\sigma$-nilpotent.

Our last result can be regarded as an extension of ^[1].

Corollary 4. Let $G$ be a $\sigma$-soluble group. Then

1. All $\sigma_i$-subgroups of $F_{\sigma}(G)$ satisfy property $\mathcal C_{\sigma_{i}}$ in $G$ for all $i \in I$.

2. $F_{\sigma}(G)$ contains every subgroup $F$ of $G$ such that, for every $i \in I$, all $\sigma_i$-subgroups of $F$ satisfy property $\mathcal C_{\sigma_{i}}$ in $G$.

Proof. Let $i \in I$ and let $A$ be a $\sigma_i$-subgroup of a $\sigma$-soluble group $G$ contained in $F_{\sigma}(G)$. Then $A$ is a $\sigma$-subnormal subgroup of $F_{\sigma}(G)$ and $F_{\sigma}(G)$ is normal in $G$, it follows that $A$ is $\sigma$-subnormal in $G$ by Lemma 1(2). Therefore $A$ satisfies property $\mathcal C_{\sigma_{i}}$ in $G$.

Assume that $F$ is a subgroup of $G$ with all its $\sigma_i$-subgroups satisfying property $\mathcal C_{\sigma_{i}}$ for every $i \in I$. By Corollary 1, every Hall $\sigma_i$-subgroup $F_i$ of $F$ is $\sigma$-subnormal in $G$. Then $F_i$ is contained in $F_{\sigma}(G)$ by ^[9]. Since $F$ is generated by its Hall $\sigma_i$-subgroups for all $i \in I$, it follows that $F \leq F_{\sigma}(G)$.

Acknowledgments

The Deanship of Scientific Research (DSR) at King Abdulaziz University (KAU), Jeddah, Saudi Arabia has funded this project, under grant no. (KEP-PhD: 20-130-1443).

Conflict of interest

The authors declare that there is no conflict of interest.